Transcript notes 13 2317 Divergence
ECE 2317 Applied Electricity and Magnetism
Spring 2014
Prof. David R. Jackson ECE Dept.
Notes 13
1
Divergence – The Physical Concept Start by considering a sphere of uniform volume charge density The electric field is calculated using Gauss's law:
z
v =
v
0
y r x a
Sphere of uniform charge density
r
<
a
:
S
Q encl
4 2
r D r D r
v
0
v
0
r
3 4 3
r
3 C/m 2 2
x
Divergence -- Physical Concept (cont.)
z r a
v =
v
0
y r
>
a
: 4
D r r
v
0
v
0 3
r
2
a
3 4 3
a
3 C/m 2 3
Divergence -- Physical Concept (cont.) Flux through a spherical surface:
S
4 2
r D r
Recall that
D r
v
0
r
3
D r
v
0 3
r
2
a
3 Hence 4 3 4 3
r a v
0
v
0 (
r
>
a
) C/m 2 C/m 2 (
r
<
a
) (increasing with distance inside sphere) (constant outside sphere) (
r
<
a
) (
r
>
a
) 4
Divergence -- Physical Concept (cont.) Observation: More flux lines are added as the radius increases (as long as we stay
inside
the charge).
S
0 The net flux out of a small volume
V
inside the charge is not zero.
V
S
S
0 Divergence is a mathematical way of describing this. 5
Gauss’s Law -- Differential Form
Definition of divergence:
div D
lim
V
0 1
V
S
V
Note: The limit exists independent of the shape of the volume (proven later).
A region with a positive divergence acts as a “source.” A region with a negative divergence acts as a “sink.” 6
Gauss’s Law -- Differential Form
Apply divergence definition to a small volume inside a region of charge:
V
S
v
(
r
)
r div D
lim
V
0 1
V
S
Gauss's law:
S
Q encl
v
Hence lim
V
0
v
1
V
v
V
V
7
Gauss’s Law -- Differential Form (cont.) The electric Gauss law in point (differential) form:
v
This is one of Maxwell’s equations .
8
x
/ 2, 0, 0
Calculation of Divergence
z div D
lim
V
0 1
S
x
z
y
(0,0,0)
x y
Assume that the point of interest is at the origin for simplicity (the center of the cube).
The integrals over the 6 faces are approximated by “sampling” at the centers of the faces.
S
D D
D y
D y D D x x z z
x
, 0, 0 2
x
, 0, 0 2 0,
y
, 0 2 0,
y
, 0 2 0, 0,
z
2 0, 0,
z
2 9
Calculation of Divergence (cont.)
div D
lim
V
0 1
S
1
S
D x
x
2 , 0, 0
D x
x
2 , 0, 0
D y
0, 2
y
, 0
D y
D D z z
0, 2
y
, 0 0, 0, 0, 0,
z
2
z
2
S
D x
D y
x
, 0, 0 2
D x
0,
y
, 0 2
x
D y
y
x
, 0, 0 2 0, 2
y
, 0
D z
0, 0,
z
2
D z z
0, 0, 2
z
10
Calculation of Divergence (cont.)
div D
lim 0
D x
D y
D z
x
, 0, 0 2
D x
x
0,
y
, 0 2
D y
y
0, 0,
z
2
D z
z
0,
x
, 0, 0 2
y
2 , 0 0, 0,
z
2 Hence
div D
D x
x
D y
y
D z
z
11
Calculation of Divergence (cont.)
Final result in rectangular coordinates:
div D
D x
x
D y
y
D z
z
12
The “del” operator
x
ˆ
x
y
z
ˆ
z
This is
a vector operator
.
Examples of
derivative operators
: scalar vector
d dx
:
x
ˆ
d dx
: Note: Any vector can be multiplied by a scalar, or multiplied by another vector in two different ways (dot and cross).
Note: The del operator is only defined in rectangular coordinates.
d dx
sin
x
cos
x
ˆ
r
ˆ
r
r
ˆ ˆ
z
ˆ
z
ˆ
x
ˆ
d
x
ˆ
dx d dx
sin
x
x
ˆ sin
x
ˆ cos
x
x
d dx
sin
x
cos
x
x
ˆ
d dx
y
ˆ sin
x
y d dx
sin
x
z
ˆ cos
x
13
Example
x
ˆ sin
x
Find
x
ˆ
x
y
ˆ
y
z
ˆ
z
x
ˆ sin
x
x
sin
x
y
z
cos
x
x
14
Divergence Expressed with del Operator Now consider:
x
ˆ
x
D x
x y
ˆ
y
z
ˆ
z
D y
y
x
D z
z y
z
Hence
D x
x
D y
y
D z
z
We therefore notice that
div D
15
Divergence with del Operator (cont.)
div D
Note that the dot
after
the del operator is important; any symbol following it tells us how it is to be used and how it is read: "gradient" "divergence" "curl" 16
Summary of Divergence Formulas
Rectangular:
D x
x
D y
y
D z
z
See Appendix A.2 in the Hayt & Buck book for a general derivation that holds in any coordinate system.
Cylindrical: 1
D
1
D
D z
z
Spherical: 1
r
2
r
2
r D r
1
r
sin
D
sin 1
r
sin
D
17
Maxwell’s Equations
(point or differential form)
H
B
t D
v
t
0 Faraday’s law Ampere’s law Electric Gauss law Magnetic Gauss law Divergence appears in two of Maxwell’s equations.
18
x
Example
Evaluate the divergence of the electric flux vector inside and outside a sphere of uniform volume charge density, and verify that the answer is what is expected from the electric Gauss law.
r z a
v =
v
0
D r
<
a
:
v
0
r
3
y
1
r
2
r
2
r D r
1
r
2
v
0
r
1
r
2 3
r
2
r
v
0
r
3 1
r
2
v
0
r
2
v
0 Note: This agrees with the electric Gauss law.
19
x
Example (cont.)
z a r
>
a
:
v =
v
0
y D r
ˆ
v
0
a
3 3
r
2 1
r
2 1
r
2 1
r
2
r
2
r D
r
r
2
r
v
0
a
3 3
r
2
v
0
a
3 3 0 0 Note: This agrees with the electric Gauss law.
20
Divergence Theorem
S n
ˆ
V V
A dV
S
In words: The volume integral of "flux per volume" equals the total flux!
21
Divergence Theorem (cont.)
Proof The volume is divided up into many small cubes.
r n
V V
A dV
lim
V
0
n N
1
A
r n
V
Note: The point
r n
is the center of cube
n.
22
Divergence Theorem (cont.)
From the definition of divergence:
r n
n
surface of cube
n
A
r n
lim 0 1
V
S n
1
V
S
n
Hence:
V
A dV
lim
V
0
n N
1
V
A
r n
lim
V
0
n N
1
V
1
V
S n
lim
V
0
n N
1
S n
23
Divergence Theorem (cont.)
r n
V V
A dV
lim 0
n N
S n
Consider two adjacent cubes: 1
n
ˆ 2
n
ˆ 1 2 ˆ is opposite on the two faces Hence, the surface integral cancels on all INTERIOR faces.
24
Divergence Theorem (cont.)
r n
V V
A dV
lim
V
0 lim
V
0
n N
S n
outside face s
S n
Hence
V
A dV
lim 0 outside f aces
S n
As 0 :
V
A dV
S
S
(proof complete) 25
x
1
z
2 Given:
Example
A
x
y
Verify the divergence theorem using this box.
S
x
x
ˆ 18 Note:
A
is constant on the front and back faces.
3
V
A dV
V
3
A x
x
x
A y
y
A z
z
3
dV
3
V
18 26
Note on Divergence Definition
div D
lim
V
0 1
V
S
Is this limit independent of the shape of the volume?
n
ˆ
r
V
S
Small arbitrary-shaped volume Use the divergence theorem for RHS:
S
V D dV div D
lim 0 1
V
lim 0 1
V
V
D dV D
r
V
D r
Hence, the limit is the same regardless of the shape of the limiting volume.
27
Gauss’s Law
(Differential to integral form) We can convert the differential form into the integral form by using the divergence theorem. Gauss’s law (differential form):
v
Integrate both sides over a volume:
V
D dV
V
v dV
Apply the divergence theorem to the LHS:
S
V
v dV
Hence
S
Q encl
28
Gauss’s Law
(Summary of two forms)
S
Definition of divergence + Gauss's law
Q encl
Integral (volume) form of Gauss’s law Divergence theorem
v
Differential (point) form of Gauss’s law Note: All of Maxwell’s equations have both a point and an integral form.
29